Solve for ( v(t) ) using initial condition (usually ( v_0 ) at ( t=0 )). The manual then often uses ( v = dx/dt ) to find ( x(t) ) with a second integration.
They forget the ( dv = -10, du ) substitution or try to integrate without separating variables first. The solutions manual shows this substitution explicitly.
Integrate both sides. The manual’s key move: substitute ( u = 2 - 0.1v ), so ( du = -0.1, dv ) → ( dv = -10, du ). [ \int \frac-10, duu = \int dt ] [ -10 \ln|u| = t + C ] [ -10 \ln|2 - 0.1v| = t + C ] Solve for ( v(t) ) using initial condition
That’s a classic variable acceleration problem. The solutions manual for Ch. 11 is correct, but let me clarify the logic.
This content is structured for different purposes: a student study guide, a blog post summary, and a Q&A for academic forums. Title: Mastering Chapter 11: Kinematics of Particles The solutions manual shows this substitution explicitly
( a = 2 - 0.1v ). And ( a = dv/dt ).
Chapter 11 of Beer & Johnston’s Vector Mechanics for Engineers: Dynamics (11th Ed.) introduces the fundamental concepts of kinematics —the geometry of motion without considering forces. This chapter is the bedrock for all future dynamics topics. [ \int \frac-10, duu = \int dt ]
Separate variables. [ \fracdv2 - 0.1v = dt ]